Evaluate the following :
`(i) int(x+2)/((x+3)(x+4)^((3)/(2))dx` `(ii) int((x+3)/(2x+5))^((1)/(2))*(1)/(x)dx`

Explanation

`(i)` Let us substitute `x+4=t^(2)` so that `dx=2tdt` and `x+2=t^(2)-2`, `x+3=t^(3)-1`.
Thus `int((x+2))/((x+3)(x+4)^(3//2))dx=int((t^(2)-2))/((t^(2)-2).t^(3))*2tdt=int(2(t^(2)-1-1))/((t^(2)-1)t^(2))dt`
`=int(2)/(t^(2))dt-2int(1)/((t^(2)-1)t^(2))dt=2int(1)/(t^(2))dt+int(2)/(t^(2))dt-2int(1)/(t^(2)-1)dt`
`=4intt^(-2)dt+int(1)/(t+1)dt-int(1)/(t-1)dt`
`=4*(t^(-1))/(-1)+ln|(t+1)/(t-1)|+C`
`=-(4)/(sqrt(x+4))+ln|(sqrt(x+4)+1)/(sqrt(x+4)-1)|+C`, `C` is a constant of integration
`(ii)` Let us substitute `(x+3)/(2x+5)=t^(2)`
so that `x=(5t^(2)-3)/(1-2t^(2))` and `dx=(-2t)/((1-2t^(2))^(2))dt`
Thus, `int((x+3)/(2x+5))^((1)/(2))*(1)/(x)dx=intt(-(1-2t^(2))/(5t^(2)-3))((-2t)/((1-2t^(2))^(2)))dt`
`=int(-2t^(2))/((5t^(2)-3)(1-2t^(2)))dt`
`=int(2t^(2))/((2t^(2)-1)(5t^(2)-3))dt`
`=6int(1)/(5t^(2)-3)dt-2int(1)/(2t^(2)-1)dt`
`=(3)/(sqrt(3).sqrt(5))ln|(sqrt(5t)-sqrt(3))/(sqrt(5t)+sqrt(3))|-(2)/(2.(1)/(sqrt(2)))ln|(sqrt(2t)-1)/(sqrt(2t)+1)|+C`
`=sqrt((3)/(5))ln|(sqrt(5)sqrt(x+3)-sqrt(2(2x+5)))/(sqrt(5(x+3))+sqrt(3(2x+5)))|-sqrt(2)ln((sqrt(2(x+3))-sqrt(2x+5))/(sqrt(2(x+3))+sqrt(2x+5)))+C`